The statement which best describes how light waves travel in an uniform medium is in straight lines. The correct answer will be A.
Answer:
7.09683 m
1.20285 s
2.4057 s
11.8 m/s
Explanation:
t = Time taken
u = Initial velocity
v = Final velocity
s = Displacement
a = Acceleration
g = Acceleration due to gravity = 9.81 m/s² (negative up, positive down)
From equation of motion we have
The maximum height above the ground that the ball reaches is 7.09683 m
Time taken to go up is 1.20285 s it will take the same time to come down so total time taken to reach the ground after it is shot is 1.20285+1.20285 = 2.4057 s
The velocity just before it hits the ground is 11.8 m/s
Answer:
62.8 μC
Explanation:
Here is the complete question
The volume electric charge density of a solid sphere is given by the following equation: ρ = (0.2 mC/m⁵)r²The variable r denotes the distance from the center of the sphere, in spherical coordinates. What is the net electric charge (in μC) of the sphere if the radius of the sphere is 0.5 m?
Solution
The total charge on the sphere Q = ∫∫∫ρdV where ρ = volume charge density = 0.2r² and dV = volume element in spherical coordinates = r²sinθdθdrdΦ
So, Q = ∫∫∫ρdV
Q = ∫∫∫ρr²sinθdθdrdΦ
Q = ∫∫∫(0.2r²)r²sinθdθdrdΦ
Q = ∫∫∫0.2r⁴sinθdθdrdΦ
We integrate from r = 0 to r = 0.5 m, θ = 0 to π and Φ = 0 to 2π
So, Q = ∫∫∫0.2r⁴sinθdθdrdΦ
Q = ∫∫∫0.2r⁴[∫sinθdθ]drdΦ
Q = ∫∫0.2r⁴[-cosθ]drdΦ
Q = ∫∫0.2r⁴-[cosπ - cos0]drdΦ
Q = ∫∫∫0.2r⁴-[-1 - 1]drdΦ
Q = ∫∫0.2r⁴-[- 2]drdΦ
Q = ∫∫0.2r⁴(2)drdΦ
Q = ∫∫0.4r⁴drdΦ
Q = ∫0.4r⁴dr∫dΦ
Q = ∫0.4r⁴dr[Φ]
Q = ∫0.4r⁴dr[2π - 0]
Q = ∫0.4r⁴dr[2π]
Q = ∫0.8πr⁴dr
Q = 0.8π∫r⁴dr
Q = 0.8π[r⁵/5]
Q = 0.8π[(0.5 m)⁵/5 - (0 m)⁵/5]
Q = 0.8π[0.125 m⁵/5 - 0 m⁵/5]
Q = 0.8π[0.025 m⁵ - 0 m⁵]
Q = 0.8π[0.025 m⁵]
Q = (0.02π mC/m⁵) m⁵
Q = 0.0628 mC
Q = 0.0628 × 10⁻³ C
Q = 62.8 × 10⁻³ × 10⁻³ C
Q = 62.8 × 10⁻⁶ C
Q = 62.8 μC
A symbiotic relationship. One where two species benefit each other. In this case, the flower gives the bee nectar, so the bee pollinates the flower.
